\(a,\dfrac{x^2-4}{2x^2y-4xy}=\dfrac{\left(x-2\right)\left(x+2\right)}{2xy\left(x-2\right)}=\dfrac{x+2}{2xy}\\ b,\dfrac{x^3+2x^2+x}{x^2+x}=\dfrac{x\left(x^2+2x+1\right)}{x\left(x+1\right)}=\dfrac{x\left(x+1\right)^2}{x\left(x+1\right)}=x+1\\ c,\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
a)
\(\dfrac{x^2-4}{2x^2y-4xy}\\ =\dfrac{\left(x-2\right)\left(x+2\right)}{2xy\left(x-2\right)}\\ =\dfrac{x+2}{2xy}\)
b)
\(\dfrac{x^3+2x^2+x}{x^2+x}\\ =\dfrac{x\left(x^2+2x+1\right)}{x\left(x+1\right)}\\ =\dfrac{\left(x+1\right)^2}{\left(x+1\right)}\\ =x+1\)
c)
\(\dfrac{7x^2+14x+7}{3x^2+3x}\\ =\dfrac{7x^2+7x+7x+7}{3x\left(x+1\right)}\\ =\dfrac{7x\left(x+1\right)+7\left(x+1\right)}{3x\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(7x+7\right)}{3x\left(x+1\right)}\\ =\dfrac{7\left(x+1\right)\left(x+1\right)}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
